2013 JAMB PHYSICS PAST QUESTIONS AND ANSWERS

$$\text{Students identify the given question paper type, marked as one of the provided options.}$$

$$\text{Mass remains constant as it is independent of location, unlike weight which depends on gravitational force.}$$

$$\text{Using vector resolution, solve for the perpendicular force using trigonometry.}$$

$$\text{Scalars have magnitude only. Volume and area qualify as they lack directional components.}$$

$$\text{Period } T \propto \sqrt{L}. \text{ If } T_1 = 2 \text{ and } L_2 = 2L_1, T_2 = 2\sqrt{2}.}$$

$$\text{Time } t = \frac{v}{a}: t = \frac{20}{2} = 10 \text{ s.}$$

$$\text{Weight apparent } W_a = mg – ma: W_a = (70)(10 – 1.5) = 595 \text{ N.}$$

`\text{Fraction immersed } = \frac{\text{Density of cork}}{\text{Density of liquid}} = \frac{0.25}{1.25} = 0.2.}

$$\text{KE = PE, } \frac{1}{2}mv^2 = mgh \to h = \frac{v^2}{2g} = \frac{5^2}{2(10)} = 1.25 \text{ m.}$$

$$\text{Power } P = \frac{Work}{time}: P = \frac{(5000)(10)(60)}{15 \times 60} = 3.3 \times 10^4 \text{ W.}$$

$$\text{Perfectly smooth surfaces lack friction, so the coefficient of friction is zero.}$$

$$\text{Effort } E = \frac{Load \times Load Arm}{Effort Arm \times Efficiency}: E = \frac{180}{2 \times 0.9} = 100 \text{ N.}$$

$$\text{Work done } W = \frac{1}{2} k x^2 \to W = F \times d: W = 20 \times 0.05 = 1.0 \text{ J.}$$

$$\text{Pressure } P = h \rho g \to h = \frac{P}{\rho g} \to h = \frac{1 \times 10^5}{1013 \times 10} \approx 10 \text{ m.}$$

$$\text{Mass of petrol } M = \rho V \to \text{Volume of alcohol } V = \frac{M}{\rho_{\text{alcohol}}} \to V = \frac{0.7 \times 4.2}{0.6} = 4.9 \text{ m³.}$$

$$\text{Length at } 20°C = \frac{20}{100} \times 400 = 80 \text{ mm.}$$

$$\text{Length change } \Delta L = L \alpha \Delta T \to L_{\text{new}} = L + \Delta L \to 100 – (100 \times 2 \times 10^{-5} \times 40) = 99.92 \text{ m.}$$

$$\text{Work } W = P \Delta V \to W = (10^5)(1.2 – 0.6) = 6 \times 10^4 \text{ J.}$$

$$\text{Since } Q = mc\Delta T, \text{ ratio of } c_Y/c_X = 2:1 \text{ as } \Delta T_X = 2\Delta T_Y.$$

$$\text{Dissolved salts elevate boiling point, allowing higher cooking temperatures.}$$

$$\text{Steam releases additional latent heat of vaporization upon contact, causing more severe burns.}$$

$$\text{Increased pressure below sea level raises boiling point above 100°C.}$$

$$\text{Surface tension allows the liquid to rise or fall, depending on its adhesive/cohesive properties with the tube material.}$$

$$\text{Conduction occurs through molecule vibration and transfer of kinetic energy in solids.}$$

$$\text{Mechanical waves require a material medium to propagate, such as tension in a string.}$$

$$\text{The main difference is the direction of vibration relative to wave propagation.}$$

$$\text{Sound velocity increases with temperature } v = v_0 + 0.6T \to v = 340 + (0.6 \times 100) = 400 \text{ m/s.}$$

$$\text{Fifth overtone is the sixth harmonic: } f_6 = 6f_1 = 6 \times 450 = 2700 \text{ Hz.}$$

$$\text{Using magnification: } \frac{Image Height}{Object Height} = \frac{Screen Distance}{Object Distance} \to h = \frac{1.5 \times 0.1}{3} = 0.05 \text{ m.}$$

$$\text{A ray passing through the center of curvature retraces its path, forming a 0° angle with the normal.}$$

$$\text{Using Snell’s law, } n_1 \sin \theta_1 = n_2 \sin \theta_2 \to \theta_1 = \arcsin(\frac{n_2 \sin \theta_2}{n_1}) \text{. Substitute values to find } \theta_1. $$

$$\text{Light must travel from a denser to a less dense medium and the angle of incidence exceeds the critical angle.}$$

$$\text{The configuration represents a telescope with inverted orientation in normal use.}$$

$$\text{Angular magnification } M = M_{\text{objective}} \times M_{\text{eyepiece}} \to M = 4 \times 7 = 28. $$

$$\text{Deviation increases with wavelength: red }<\text{ green }<\text{ blue. }$$

$$\text{Using } F = qE, \text{ substitute } q = 1.5 \times 10^{-19}, E = 10^5 \text{ to find } F = 1.5 \times 10^{-14} \text{ N.} $$

$$\text{Capacitors help prevent electric sparks by absorbing voltage spikes in circuits.}$$

$$\text{Using } V = IR + Ir, \text{ rearrange to find internal resistance: } r = \frac{V – IR}{I} \to r = 1.0 \Omega. $$

$$\text{Ohm’s law applies to metals and other materials with consistent resistance under constant conditions.}$$

$$\text{Total power = (10 \times 40 + 5 \times 100) W, time = 10 h. Calculate energy (kWh) and multiply by cost per unit.}$$

$$\text{Using } I = \frac{P}{V}, \text{ substitute } P = 2000 \text{ W and } V = 250 \text{ V to find } I = 8 \text{ A.} $$

$$\text{Charge gains kinetic energy while losing potential energy due to work done against the field.}$$

$$\text{A high resistance connected in series with the galvanometer increases its range to measure voltage.}$$

$$\text{Faraday’s law states that induced e.m.f. is proportional to the rate of change of magnetic flux.}$$

$$\text{Mass deposited is proportional to current and time: } m = \frac{I_t}{I_0 t_0} \times m_0 \to m = 4.6 \text{ g.}$$

$$\text{Energy at level } n = \frac{E_1}{n^2} \to E = \frac{-24}{3^2} = -2.75 \text{ eV.} $$

$$\text{The number of photoelectrons is proportional to the intensity of the light beam incident on the surface.}$$

$$\text{Photoelectric effect demonstrates light’s particle nature as energy is transferred in discrete photons.}$$

$$\text{Using } E = \frac{hc}{\lambda}, \text{ substitute values to find } E = 2.0 \times 10^{-15} \text{ J.} $$

$$\text{Covalent bonding forms due to shared electrons between silicon and germanium.}$$